Explanation

(b)
Let $y = 4e^{2x} + 9e^{- 2x}\ \ \Rightarrow \ \ \ \ \ \frac{dy}{dx} = 8e^{2x} - 18e^{- 2x}$
$\Rightarrow \ \ \ \ \frac{d^{2}y}{dx^{2}} = 16e^{2x} + 36e^{- 2x}$
For minimum, put $\frac{dy}{dx} = 0\ \ \ \Rightarrow \ \ 8e^{2x} - 18e^{- 2x} = 0$
$\Rightarrow \ \ \ \ e^{4x} = \frac{9}{4}\ \ \ \Rightarrow \ \ \ \ x = \frac{1}{4}\ \log\frac{9}{4}$
At $x = \frac{1}{4}\ \log{\frac{9}{4},\ \ \ \frac{d^{2}y}{dx^{2}} > 0}$
$\therefore$ Minimum value at $x = \frac{1}{4}\log\frac{9}{4}$ is
$y = 4e^{2\left( \frac{1}{4}\log\frac{9}{4} \right)} + 9e^{- 2\left( \frac{1}{4}\log\frac{9}{4} \right)}$
$= 4e^{\log\left( \frac{9}{4} \right)^{1/2}} + 9e^{\log\left( \frac{9}{4} \right)^{- 1/2}}$
$= 4.\frac{3}{2} + 9.\frac{2}{3} = 12\ \$